Problem 25
Question
Find the fifth term and the nth term of the geometric sequence whose first term \(a_{1}\) and common ratio \(r\) are given. $$ a_{1}=\sqrt{3} ; \quad r=\sqrt{3} $$
Step-by-Step Solution
Verified Answer
The fifth term is \(9\sqrt{3}\). The nth term is \(a_n = \sqrt{3} \times 3^{(n-1)/2}\).
1Step 1: Formula for the nth term of a geometric sequence
The nth term of a geometric sequence can be found using the formula: \[ a_n = a_1 \times r^{(n-1)} \] where \(a_n\) is the nth term, \(a_1\) is the first term, \(r\) is the common ratio, and \(n\) is the term number.
2Step 2: Find the fifth term (\(a_5\))
To find the fifth term, substitute \(a_1 = \sqrt{3}\), \(r = \sqrt{3}\), and \(n = 5\) into the nth term formula:\[ a_5 = \sqrt{3} \times (\sqrt{3})^{(5-1)} \]\[ a_5 = \sqrt{3} \times (\sqrt{3})^4 \]Simplify \((\sqrt{3})^4\):\[ (\sqrt{3})^4 = 3^2 = 9 \]Now multiply:\[ a_5 = \sqrt{3} \times 9 = 9\sqrt{3} \]
3Step 3: General formula for the nth term
Substitute the given values \(a_1 = \sqrt{3}\) and \(r = \sqrt{3}\) into the general formula:\[ a_n = \sqrt{3} \times (\sqrt{3})^{(n-1)} \] Simplify the exponent:\[ a_n = \sqrt{3} \times 3^{(n-1)/2} \]So the nth term formula for this geometric series is:\[ a_n = \sqrt{3} \times 3^{(n-1)/2} \]
Key Concepts
nth term formulacommon ratiogeometric series
nth term formula
To understand geometric sequences, we need to start with the nth term formula. This formula helps us find the term at any position in a geometric sequence. The nth term formula is given by:
\( a_n = a_1 \times r^{(n-1)} \).
Here:
In our problem, the first term \( a_1 \) is \( \sqrt{3} \) and the common ratio \( r \) is also \( \sqrt{3} \). By substituting these values into the formula, you can find any term in the sequence.
\( a_n = a_1 \times r^{(n-1)} \).
Here:
- \( a_n \) is the term you are trying to find,
- \( a_1 \) is the first term of the sequence,
- \( r \) is the common ratio, and
- \( n \) is the term number.
In our problem, the first term \( a_1 \) is \( \sqrt{3} \) and the common ratio \( r \) is also \( \sqrt{3} \). By substituting these values into the formula, you can find any term in the sequence.
common ratio
The common ratio, \( r \), is a key part of any geometric sequence. It is the factor by which you multiply each term to get the next term. If you know the common ratio and the first term, you can find all terms in the sequence.
In this exercise, \( r = \sqrt{3} \). So, if our first term \( a_1 \) is \( \sqrt{3} \), the second term will be \( \sqrt{3} \times \sqrt{3} = 3 \).
The third term will be \( 3 \times \sqrt{3} = 3\sqrt{3} \), and so on. The common ratio remains consistent throughout, and it dictates the pattern of the sequence.
Once you understand the common ratio, predicting the next terms becomes straightforward.
In this exercise, \( r = \sqrt{3} \). So, if our first term \( a_1 \) is \( \sqrt{3} \), the second term will be \( \sqrt{3} \times \sqrt{3} = 3 \).
The third term will be \( 3 \times \sqrt{3} = 3\sqrt{3} \), and so on. The common ratio remains consistent throughout, and it dictates the pattern of the sequence.
Once you understand the common ratio, predicting the next terms becomes straightforward.
geometric series
A geometric series is the sum of the terms in a geometric sequence. If we were to sum the first few terms of our sequence, we would be calculating a geometric series.
For instance, with the first few terms being \( \sqrt{3}, 3, 3\sqrt{3} \), the sum would be \( \sqrt{3} + 3 + 3\sqrt{3} \).
Finding the sum of a geometric series, especially if it has many terms, involves specific formulas. The sum to the nth term, \( S_n \), can be given by: \[ S_n = a_1 \frac{1-r^n}{1-r} \]\ if \( r \eq 1 \).
Using this formula makes it easy to find the sum of multiple terms without adding one by one. Mastering this concept is essential for tackling more complex mathematical problems involving sequences.
For instance, with the first few terms being \( \sqrt{3}, 3, 3\sqrt{3} \), the sum would be \( \sqrt{3} + 3 + 3\sqrt{3} \).
Finding the sum of a geometric series, especially if it has many terms, involves specific formulas. The sum to the nth term, \( S_n \), can be given by: \[ S_n = a_1 \frac{1-r^n}{1-r} \]\ if \( r \eq 1 \).
Using this formula makes it easy to find the sum of multiple terms without adding one by one. Mastering this concept is essential for tackling more complex mathematical problems involving sequences.
Other exercises in this chapter
Problem 25
Prove each statement. $$ \begin{aligned} &a-b \text { is a factor of } a^{n}-b^{n}\\\ &\text { [Hint: } \left.a^{k+1}-b^{k+1}=a\left(a^{k}-b^{k}\right)+b^{k}(a-
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Expand each expression using the Binomial Theorem. $$ (\sqrt{x}+\sqrt{2})^{6} $$
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In Problems 25-30, find the indicated term in each arithmetic sequence. $$ \text { 100th term of } 2,4,6, \ldots $$
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List the first five terms of each sequence. \(\left\\{b_{n}\right\\}=\left\\{\frac{n}{e^{n}}\right\\}\)
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